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kow [346]
3 years ago
15

What is equivalent to log 25 125

Mathematics
1 answer:
nexus9112 [7]3 years ago
5 0
Log25 (5•25) = 3 over 2 aka 3/5
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Suppose that diameters of a new species of apple have a bell-shaped distribution with a mean of 7.25cm and a standard deviation
Ulleksa [173]

Answer:

95%

Step-by-step explanation:

5 0
3 years ago
If f(x)=3x^2-5 and g(x)=-2x+5, then f(3)g(3)=
Vladimir [108]

Answer:

(

f

+

g

)

(

x

)

=

3

x

2

+

x

+

2

Explanation:

(

f

+

g

)

(

x

)

=

f

(

x

)

+

g

(

x

)

:

(

f

+

g

)

(

x

)

=

3

x

2

−

x

+

5

+

2

x

−

3

Add like terms:

(

f

+

g

)

(

x

)

=

3

x

2

+

(

−

x

+

2

x

)

+

(

5

−

3

)

(

f

+

g

)

(

x

)

=

3

x

2

+

x

+

2

(

f

+

g

)

(

x

)

=

3

x

2

+

x

+

2

Explanation:

(

f

+

g

)

(

x

)

=

f

(

x

)

+

g

(

x

)

:

(

f

+

g

)

(

x

)

=

3

x

2

−

x

+

5

+

2

x

−

3

Add like terms:

(

f

+

g

)

(

x

)

=

3

x

2

+

(

−

x

+

2

x

)

+

(

5

−

3

)

(

f

+

g

)

(

x

)

=

3

x

2

+

x

+

2

(

f

+

g

)

(

x

)

=

3

x

2

+

x

+

2

Explanation:

(

f

+

g

)

(

x

)

=

f

(

x

)

+

g

(

x

)

:

(

f

+

g

)

(

x

)

=

3

x

2

−

x

+

5

+

2

x

−

3

Add like terms:

(

f

+

g

)

(

x

)

=

3

x

2

+

(

−

x

+

2

x

)

+

(

5

−

3

)

(

f

+

g

)

(

x

)

=

3

x

2

+

x

+

2

Step-by-step explanation:

5 0
3 years ago
Pls help show work please
Aloiza [94]
Each bracelet is $5, this can be represented as 5b. (Where b is number of bracelets)

Each necklace is $8, this can be represented as 8n. (Where n is number of necklaces)

We know that the total money raised needs to be at least $100, but it can also be more than that.

So we know that 8n+5b will be greater than or equal too 100.

8n+5b ≥ 100
3 0
3 years ago
What is 3 to the 6th power to the negative 2
malfutka [58]

I believe the answer is 1/531441

8 0
3 years ago
Assume the competing hypotheses take the following form: H0: µ1 – µ2 = 0, HA: µ1 – µ2 ≠ 0, where µ1 is the population mean for p
DedPeter [7]

Answer:

t=\frac{\bar x_1- \bar x_2}{\sqrt{\frac{s_1^2}{n_1} +\frac{s_2^2}{n_2} } }

Step-by-step explanation:

H0: µ1 – µ2 = 0

HA: µ1 – µ2 ≠ 0

We have given,

The population variances are not known and cannot be assumed equal.

The test statistic for the test is

t=\frac{\bar x_1- \bar x_2}{\sqrt{\frac{s_1^2}{n_1} +\frac{s_2^2}{n_2} } }

Where,

\bar x_1 = sample meaan of population 1

\bar x_2 = sample mean of population 2

n_1 = sample size of population 1

n_2 = sample size of population 2

Therefore, this is the test

t=\frac{\bar x_1- \bar x_2}{\sqrt{\frac{s_1^2}{n_1} +\frac{s_2^2}{n_2} } }

7 0
3 years ago
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